Mixing Problem with Initial Condition
Tags | |
UUID | 534059c8-395a-11e6-9770-bc764e2038f2 |
In differential equations, mixing problems are used to model concentrations of a substance dissolved in a fluid. This can include anything from salt content in water to pollution in air. When doing mixing problems, we make the following assumptions:
- Fluid flows into the mixture at some rate, and with a certain concentration of the substance (possibly 0).
- Fluid flows out of the mixture at some rate, and with some concentration of the substance.
- The substance and fluid are well mixed so that the concentration is uniform throughout the volume.
For this particular equation, we make the further assumption that the rate at which fluid flows into the mixture is the same as the rate at which fluid flows out so that the volume is constant, but this will not always be the case. We also assume the concentration of the substance in the fluid flowing in and the flow rates are constant and not a function of time, though again this will not always be the case.
Derivation of Solution
This basic version of the mixing problem produces a first order linear differential equation. We have dSdt=rate in - rate out, where dSdt is the rate of change of the concentration of the substance in the mixture. The rate in is (rate of fluid flowing in)⋅(concentration of substance in the fluid flowing in), and the rate out is (rate of fluid flowing out)(volume of mixture)⋅S(t), where S(t) is the concentration of the substance in the mixture at time t. And again, in this version of the problem we're assuming that the rate flowing in is equal to the rate flowing out. We'll call this rate R, the concentration C, and the volume V.
So
dSdt=R⋅C-R⋅S(t)V
This is our first order linear differential equation. To solve it we use the method of integrating factors. Moving the rate out term to the other side, we have
dSdt+R⋅S(t)V=R⋅C
So we want to find e∫R/Vdt. Since neither R nor V depend on time, we know our integrating factor is eRt/V. We multiply this through both sides to get
eRt/V⋅dSdt+RV⋅eRt/V⋅S(t)=eRt/V⋅R⋅C, which is equivalent to
(eRt/V⋅S(t))′=eRt/V⋅R⋅C
Integrating both sides with respect to t gives
∫(eRt/V⋅S(t))′=∫eRt/V⋅R⋅Cdt
eRt/V⋅S(t)=VR⋅R⋅C⋅eRt/V+K
S(t)=V⋅C+K⋅e-(Rt/V)
Where K is the constant of integration determined by the initial conditions.
Example with Initial Condition
Suppose you have a tank which holds 200 gallons of pure water. At time t=0 a solution of 3lb of salt per gallon flows into the tank at a rate of 4 gallons per minute. At the same time, a well-mixed solution begins to flow out of the tank at a rate of 4 gallons per minute.
To set up the equation, we need to determine what our rate in and rate out are.
rate in=(rate of fluid flowing in)⋅(concentration of salt in the fluid flowing in)=3⋅4=12
rate out=(rate of fluid flowing out)(volume of mixture)⋅S(t)=4⋅S(t)200
So our differential equation is
dSdt=12-4⋅S(t)200,
and we want to find an equation for S(t).
We move the rate out part to the left hand side and get
dSdt+4⋅S(t)200=12
and find the integrating factor:
e∫4/200dt=e4t/200.
We multiply this through both sides to get
e4t/200dSdt+4200e4t/200S(t)=12e4t/200
So by the method of integrating factors, we get to
[e4t/200S(t)]′=12e4t/200
We integrate both sides with respect to t to get
e4t/200S(t)=∫12e4t/200
e4t/200S(t)=600e4t/200+C
S(t)=600+Ce-4t/200
This is the general solution for the differential equation. To find the specific solution and find the value of C we need to use our initial conditions. We know that before the salt solution started to enter the tank it was pure water, so in terms of our equations that means at t=0 we had S(0)=0. Plugging this in,
0=600+Ce0
-600=C.
So with our initial conditions applied, we can write the final solution:
S(t)=600-600e-4t/200
Reference
Polking, John C., Albert Boggess, and David Arnold. "Mixing Problems." Differential Equations with Boundary Value Problems. Upper Saddle River, NJ: Pearson/Prentice Hall, 2006. N. pag. Print.
- Comments
- Attachments
- Stats
No comments |